The thief $T$ can escape if $C$ is a circle, with a simple strategy of dribbling left and right each policeman at a time in such a way that he is left out of reach of the thief no matter what the future dribbles will be.
The key insight is due to Pietro Majer: the thief can approach $C$ in such a way that its shadow $S$ (closest point) on $C$ moves at speed faster than $1$.
A few trivial facts and clarifications are needed to describe the strategy:
Assume $C$ has radius $1$, then pick a number $r$ once and for all, with $1/2<r<1$. It's obvious that wherever $T$ may be, there are exactly two circles of radius $r$, tangent to $C$ and contaning $T$:
As $T$ moves along one or the other of the two circles towards the landing points ($A$ or $B$), the second circle is not static but dragged along with $T$, so that $T$ is always the intersection point.
No matter how often $T$ zigzags (sometimes moving towards $A$, sometimes towards $B$) the path followed by $T$ has the same length, since $A$ and $B$ always have equal distances $\overset{\frown}{TA}=\overset{\frown}{TB}$ from $T$. At the end $T$ may land anywhere between $A$ and $B$, but in finite time, independent of the turns.
The fact that $r>1/2$ guarantees that $S$ moves at speed $>1$, that is $\overset{\frown}{SA}>\overset{\frown}{TA}$. This is a tedious but easy trigonometric inequality, better left to the reader.
Any policeman to the right of $A$, or to the left of $B$, by more than $\overset{\frown}{SA}$, is inactive, he will never catch $T$.
Finally the strategy: enumerate the policemen $P_1, P_2, \dots$. Take the first active policeman, say $P$. If $P$ is to the left of $S$ (or on $S$) $T$ will move to the right towards $A$ at speed $1$. (Similarly if $P$ is to the right, $T$ will move left towards $B$.) Since $S$ moves at speed $>1$ at some point $P$ will fall and remain behind $S$ by $\epsilon$ (it doesn't matter how small). But $P$ may still be active so $T$ continues to move towards $A$ until $\overset{\frown}{SA}<\epsilon/3$. That makes $P$ inactive: he is now even behind $B$ by more than $\epsilon/3$!
Next we take the next active policeman $Q$ on the list and again $T$ moves in the direction away from him, until $Q$ is inactive too. There are only countably many policemen and they will all eventually become inactive, guaranteeing that $T$ lands on a police-free point of $C$.
This proof easily extend to any curve $C$, since $T$ can first move close to a point of positive curvature, where locally the curve can be approximated well by a circle.